Find frac{dy}{dx} if y = sec^{-1}left(frac{1} | vedclass

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(B) Given $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$.Using the identity $\sec^{-1}(u) = \cos^{-1}(1/u)$,we have:$y = \cos^{-1}(2x^2 - 1)$.Let $x =

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$\frac{-2}{\sqrt{1-x^2}}$

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(B) Given $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}ight)$.Using the identity $\sec^{-1}(u) = \cos^{-1}(1/u)$,we have:$y = \cos^{-1}(2x^2 - 1)$.Let $x = \cos 	heta$,then $2x^2 - 1 = 2\cos^2 	heta - 1 = \cos(2	heta)$.Since $0 Thus,$y = \cos^{-1}(\cos 2	heta) = 2	heta = 2\cos^{-1}x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \left(\frac{-1}{\sqrt{1-x^2}}ight) = \frac{-2}{\sqrt{1-x^2}}$.

Find $\frac{dy}{dx}$ if $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$,where $0 < x < \frac{1}{\sqrt{2}}$.

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